Prascius 1Steven PrasciusMrs. Tallman AP Calculus17 March 2014Riemann Sums, Trapezoid Rule, and Simpsons RuleWhen determining the area under the curve of a function, the definite integral is almost always used. Although the definite integral is by far the most accurate method for determining the area under a curve, there are many other methods. These other methods are Riemann sums, the trapezoid rule, and Simpsons rule.One method for finding the area under the curve is Riemann sums. A Riemann sum approximates the area under the curve by separating the area under the curve into separate rectangles and adding the areas of the rectangles together. The formal notation of a Riemann sum is Rn= i=1nfx*x where x =(b-a)/n. In this notation n is the number of intervals, or number of rectangle that will be used, a and b are the start and end points for the integral, and f(x) is the height of the rectangle, or y-value. This notation simply means that the areas of the rectangles, found by multiplying height (f(x) by width (x), are added together to get an estimate for the area under the curve. There are five different types of Riemann sums which are left, right, midpoint, upper, and lower Riemann sums. All of these types of Riemann sums are found using rectangles with the only difference being the height of the rectangles they use. All of the types of Riemann sums use the same interval and same widths of the rectangles but use different heights found at different parts of the integrals. Left Riemann sums use heights, or f(x) values, that are found on the left most portion of each interval while right Riemann sums use heights that are found on the right most portion of each interval. Midpoint Riemann sums use the height value found directly in the center of the interval. Upper Riemann sums use the heights that correspond to the highest f(x) value in the integral and lower Riemann sums use heights that correspond to the lowest f(x) value in the integral. The differences between the types of Riemann sums can be seen in figure 1 below. Midpoint Riemann SumRight Riemann SumLeft Riemann SumLower Riemann SumUpper Riemann SumFigure 1. Types of Riemann SumsAs seen in the figure above, different types of Riemann sums over or under estimate the area under the curve depending on the shape of the function. Because of this it is important to choose a type of Riemann sum that will most closely fit the area under the curve for a given function to get an accurate estimate of area.An application of the five types Riemann sums can be seen when considering the function given by f(x) = (x-3)4 + 2(x-3)4 - 4(x-3) + 5 in the interval from x=1 to x=5. If only two rectangles are used for each Riemann sum, the intervals for each would be x=1 to x=3 and x=3 to x=5. All of the Riemann sums have the same width of two because each interval is two units wide but would have different heights. If we were to calculate the left and right Riemann sums first, the left Riemann sum would use f(1) and f(3) for the height of its rectangles because those are the left most parts of the interval and the right Riemann sum would use f(3) and f(5) for its heights because those are the right most parts of the interval. A visual representation of the right and left Riemann sums can be seen in figure 2 below.Right Riemann SumLeft Riemann SumFigure 2. Left and Right Riemann SumsWhen the Riemann sums are calculated they come out to be:Left2 = f(1)*2 + f(3) * 2Right2 = f(3) *2 + f(5) * 2Left2 = 13*2 + 5*2Right2 = 5*2 + 29 * 2Left2 =36 units2Right2 = 68 units2Both the left and right Riemann sums overestimate the actual area under the curve because both have rectangles that encompass more area than just the area under the curve (shown in red). Given the same information the midpoint Riemann sum would use two as the width of its rectangles and use the midpoints of the intervals as its heights which are f(2) and f(4). A visual representation of this Riemann sum can be seen in figure 3.Midpoint Riemann SumFigure 3. Midpoint Riemann SumsWhen calculated the Midpoint Riemann Sum would come out to be:Midpoint2 = f(2)*2 + f(4) *2Midpoint2 = 8*2 + 4*2Midpoint2 = 24 units2The midpoint Riemann sum for this problem underestimates the area under the curve because its rectangles leave out much of the area under the curve. The upper and lower Riemann sums for this example would use heights that are a combination of those used in the previous Riemann sums. The upper Riemann sum would have widths of two for its rectangles and heights of f(1) and f(5) because those are points on each interval where f(x) has its highest values. The lower Riemann sum would have two as the width of its rectangles and heights of f(3) and f(3.7) because those are the points on the intervals where f(x) has its lowest values. A visual representation of these Riemann sums can be seen in figure 4 bel